A spring of spring constant k=8.25N/m is displaced from equilibrium by a distance of 0.150m. What is the stored energy in the form of spring potential energy? Show your work.

it's a form of government where people elect their representatives

Answer:

The word democracy itself means rule by the people.

According to the Newton's second law :- The acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

d. An object did work on the environment.

Work is defined in many contexts. Some of these are;

i. Work is the product of force and displacement. In this case, work done is positive if the force applied on an object or body and the displacement caused by the force are in the same direction. If instead the force and displacement are in opposite direction, then the work done will be negative. If it is the case the force and the displacement are perpendicular to each other, the work done is zero.

ii. In the first law of thermodynamics, the internal energy of a system is the sum of the work done and the heat exchanged between the system and the environment. Therefore, work done is the difference between the internal energy of a system and the heat exchanged between the system and the environment.

In this case, work is said to be positive if work is done by the system (object) on the environment. It is negative if work is done by the environment on the system (object).

Answer:

its c

Explanation:

Answer:

inertia

Explanation:

Answer:

The velocity of the fish when it hits the water is:

v = 10.93 m/s and 71.88 ° below the x-direction.

Explanation:

Let's find the velocity of the fish in the y-direction.

[tex]v_{fy}^{2}=v_{iy}^{2}-2gh[/tex]

Here, v(iy) of the fish is zero, and the heigh h = 5.50 m, then the velocity will be:

[tex]v_{fy}^{2}=0-2(9.81)(5.50)[/tex]

[tex]v_{fy}^{2}=-2(9.81)(5.50)[/tex]

[tex]v_{fy}=-10.39 \: m/s[/tex]

Now, we know that the velocity in the x-direction is constant, so we can calculate the velocity of the fish when it hits the water.

[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]

[tex]v=\sqrt{3.40^{2}+(-10.39)^{2}}[/tex]

[tex]v=10.93 \: m/s[/tex]

And the direction will be:

[tex]\theta=tan^{-1}(\frac{10.39}{3.40})[/tex]

[tex]\theta=71.88^{\circ}[/tex]

The angle is 71.88 ° belox the x-direction.

I hope it helps you!

Answer:

the period of oscillation of the given object is 0.14 s

Explanation:

Given;

mass of the object, m = 3 kg

extension of the spring, x = 0.085 m

The spring constant is calculated as follows;

[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]

The angular speed of a 4 kg object is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]

Therefore, the period of oscillation of the given object is 0.14 s

Answer:

the rms value of the electric field component transmitted is 3.295 V/m

Explanation:

Given;

intensity of the unpolarized light, I = 0.0288 W/m²

For unpolarized light, the relationship between the amplitude electric field and intensity is given as;

[tex]E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m[/tex]

The relationship between the rms value of the electric field and the amplitude electric field is given as;

[tex]E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m[/tex]

Therefore, the rms value of the electric field component transmitted is 3.295 V/m

Answer:

 v₂ /v₁ = 2.3 10⁺²

Explanation:

The energy is conserved so the total potential energy must be transformed into kinetic energy

          K = U

          ½ m v² = q ΔV

          v = [tex]\sqrt{\frac{2q \Delta V}{m} }[/tex]

a) Let's find the speed of the electron

           m = 9.1 10⁻³¹ kg

as they do not indicate the value of the power difference, we will assume that ΔV = 1 V is worth one

           v = [tex]\sqrt{ \frac{2 \ 1.6 \ 10^{-19} \ 1}{9.1 \ 10^{-31}} }[/tex]

           v = [tex]\sqrt {0.3516 \ 10^{12}}[/tex]

           v1 = 0.593 10⁶ m / s

b) the velocity of a hydrogen ion

            M = M_H + m

            M = 1.673 10⁻²⁷ + 9.1 10⁻³¹

            M = 1.67391 10⁻²⁷ kg

            M = 1.67 10⁻²⁷ kg

            v = [tex]\sqrt{ \frac { 2 \ 1.6 \ 10^{-19} \ 1}{1.67 \ 10^{-27}} }[/tex]

             v = [tex]\sqrt{ 1.916167 \ 10^8 }[/tex]

             v₂ = 1.38 10⁴ m / s

the relationship between these speeds is

           v₂ / v₁ = 1.38 10⁴ / 0.593 10⁶

           v₂ /v₁ = 2.3 10⁺²

Answer:

[tex]T=22.5sec[/tex]

Explanation:

From the question we are told that:

Mass of astronaut [tex]m_a=105kg[/tex]

Mass of tool [tex]m_t=16kg[/tex]

Distance [tex]d=18m[/tex]

Velocity of separation [tex]v_s= 0.1m/s[/tex]

Velocity of tool bag [tex]v_t=4.5m/s[/tex]

Generally the equation for momentum is mathematically given by

 [tex]P=mv[/tex]

Therefore

Initial Momentum before drop

 [tex]P_1=0.1(105+16)[/tex]

 [tex]P_1=12.1[/tex]

Initial Momentum after drop

 [tex]P_2=-16(4.5)+105V[/tex]

Therefore

Since [tex]P_1=P_2[/tex]

 [tex]-72+105V=12.1[/tex]

 [tex]V=0.8m/s[/tex]

Generally the equation for Time T is mathematically given by

 [tex]T=\frac{d}{V}[/tex]

 [tex]T=\frac{18}{0.8}[/tex]

 [tex]T=22.5sec[/tex]

Answer:

Moving electrons always create a magnetic field. Electrons moving along a wire make a magnetic field that goes in circles around the wire. When you bend the wire into a coil, the magnetic fields around each loop of the coil add up to make a long , thin magnet with north at one end and south at the other.

Explanation:

Answer:

noun

the speed of something in a given direction.

"the velocities of the emitted particles"

(in general use) speed.

"the tank shot backwards at an incredible velocity"

Similar:

speed

pace

rate

tempo

momentum

impetus

swiftness

swift/fast pace

fastness

quickness

speediness

rapidity

briskness

expeditiousness

expedition

dispatch

acceleration

clip

fair old rate

fair lick

steam

nippiness

fleetness

celerity

ECONOMICS

the rate at which money changes hands within an economy.

noun: velocity of circulation; plural noun: velocities of circulation

Answer:

i hope this helps you

Explanation:

hii

Answer:

-2.869 rad/s2

Explanation:

Data given:

speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)

Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds

And for every 1km, there will be 1000m.

So, speed of 95.0 km/h = 26.389 m/s

speed, vi  =  r ω (radius X angular velocity)

 angular velocity, ωi = v/r

ωi = 26.389  m/s ÷ half of 0.88 m diameter

= 59.975 rad/s

decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)

= 16.667m/s

The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter

 = 37.879rad/s

 How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ

= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)

= -2.869 rad/s2

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

Answer:

The intensity of light passing from the third polarizer is 3Io/16.

Explanation:

The law of Malus is given by

[tex]I=I_o cos^2\theta[/tex]

Let the incident intensity of light is Io.

The intensity of light passing from the first polarizer is

[tex]I' = \frac{I_o}{2}[/tex]

The intensity of light passing from the second polarizer is

[tex]I''=\frac{I_o}{2}\times cos^230 =\frac{3I_o}{8}[/tex]

The intensity of light passing from the third polarizer is

[tex]I''' = \frac{3I_o}{8}\times cos^2 60\\\\\\I''' = \frac{3I_o}{16}[/tex]

Answer:

Copper would experience induced magnestism mostly easily.

Explanation:

Permalloy would experience induced magnetism most easily.

I don't know the options but that is correct too!!

True

Hope this helps! :)

Answer:

true because when trow the frisbee gives u level

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

Answer:

[tex]\lambda=5.46\times 10^{-8}\ m[/tex]

Explanation:

The hottest ordinary star in our galaxy has a surface temperature of 53,000 K.

We need to find the peak wavelength of its thermal radiation.

Using Wein's law,

[tex]\lambda T=2.898\times 10^{-3}\\\\\lambda=\dfrac{2.898\times 10^{-3}}{53000}\\\\=5.46\times 10^{-8}\ m[/tex]

So, the peak wavelength of its thermal radiation is equal to [tex]5.46\times 10^{-8}\ m[/tex].

Answer:

The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]

Explanation:

This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:

[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)

Where:

[tex]m[/tex] - Mass of the car, in kilogram.

[tex]v[/tex] - Initial velocity, in meters per second.

[tex]\mu[/tex] - Coefficient of friction, no unit.

[tex]s[/tex] - Travelled distance, in meters.

Then we derive an expression for the distance travelled by the vehicle:

[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]

[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]

As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]

Answer:

The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.

Explanation:

Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."

Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.

Answer:

The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.

Sample Problem #1

A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 900 kg

v = 10.0 m/s

R = 25.0 m

Requested Information:

a = ????

Fnet = ????

To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (10.0 m/s)2 / (25.0 m)

a = (100 m2/s2) / (25.0 m)

a = 4 m/s2

To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.

Fnet = m • a

Fnet = (900 kg) • (4 m/s2)

Fnet = 3600 N

Sample Problem #2

A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 95.0 kg

R = 12.0 m

Traveled 1/4-th of the circumference in 2.1 s

Requested Information:

v = ????

a = ????

Fnet = ????

To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:

v = d / t

v = (0.25 • 2 • pi • R) / t

v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)

v = 8.97 m/s

To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (8.97 m/s)2 / (12.0 m)

a = (80.5 m2/s2) / (12.0 m)

a = 6.71 m/s2

To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.

Fnet = m*a

Fnet = (95.0 kg)*(6.71 m/s2)

Fnet = 637 N

In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.

Answer:

[tex]d =3.7*10^{-3} m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.2kg[/tex]

Spring constant [tex]\mu=22Nm^{-1}[/tex]

Amplitude [tex]A=5cm=0.05m[/tex]

Generally the equation for displacement d is mathematically given by

 [tex]d = Asin(\omega t)[/tex]

Where

 [tex]\omega=angular\ velocity[/tex]

 [tex]\omega=\sqrt{k/m}[/tex]

 [tex]\omega=\sqrt{22/1.2}[/tex]

 [tex]\omega=4.2817rads^{-1}[/tex]

Therefore  

 [tex]d = 0.05*sin(4.2817*1)[/tex]  

 [tex]d =3.7*10^{-3} m[/tex]

Answer:

Tx not but mybe

Explanation:

for that reason its just trying to help

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

Given the following data;

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

[tex] Resistance = P \frac {L}{A} [/tex]

Where;

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]

[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]

Resistance = 9.95 Ohms

Answer:

According to " The second law of thermodynamics", the randomness of the universe is constantly increasing?

Explanation:

So answer option C. Have a great summer.

Answer:

By the Pythagorean Theorem the distances from the speakers os

5 and 5.5 (rounding) meters   - let y be the wavelength in the solution

n y = 5      n is number of wavelengths from speaker

(n + m) y = 5.5      m must be integral for constructive interference

m y = .5       subtracting equations

m = 2      and y = ,25   for the above conditions

(n + 2) y = 5.5     substituting for m

n = 5.5 / .25 - 2 = 20

f = v / y     using frequency of sound

f = 340 / .25 = 1360 / sec    for lowest frequency

Check: D1 = y n  = ,25 * 20 = 5

and D2 = .25 * 22 = 5.5     for the distances traveled